Question

Which of the following gas has the rate of diffusion 0.88 times that of Phosphine (PH3) ?

a) NO2 b) N2O c) CH4 d) C2H2

Answer 1

`"b) N"_2"O"`

Explanation:

Rate of diffusion `("r")` and molar mass `("M")` are ralated as

`"r" ∝ 1/sqrt"M"`

`"r"_("PH"_3)/"r" = sqrt("M"/"M"_("PH"_3))`

`cancel("r"_("PH"_3))/(0.88 cancel("r"_("PH"_3))) = sqrt("M"/"34 g/mol")`

`1/0.88 = sqrt("M"/"34 g/mol")`

`"M" = "34 g/mol"/0.88^2 ≈ "44 g/mol"`

`underline(bb("Molecule") color(white)(......)bb("Molar mass"))`
`"NO"_2 color(white)(................)"46.0 g/mol"`
`"N"_2"O" color(white)(................)"44.0 g/mol"`
`"CH"_4 color(white)(................)"16.04 g/mol"`
`"C"_2"H"_2 color(white)(..............)"26.04 g/mol"`

∴ `"b) N"_2"O"` is correct answer.